- Each of these triangles has a point U = midpoint of GH. The Fermat

line goes through this point, so either the U points of the three

triangles involved are colinear, or they are the same.

Steve

On Dec 7, 2005, at 9:22 AM, Wilson Stothers wrote:

> Dear All,

>

> We have the following fact :

>

> If F is a Fermat Point of triangle ABC, then F is also a

> Fermat Point of its Cevian, and of its Anticevian Triangle.

>

> A couple of questions :

>

> Is this well-known? I'd appreciate a reference.

>

> Apart from G, are there other centres with this property?

>

> I also observe from Cabri that, if F is a Fermat point,

> and if ABC and the Cevian and Anticevian Triangles have

> Fermat Points {F,F2},{F,F3},{F,F4}, respectively, then

> (a) the points F,F2,F3,F4 are concyclic,

> (b) the circle also contains the isogonal conjugates of

> F2, F3, F4 with respect to their triangles,

> (c) the point F and its isogonal conjugates in the three

> triangles are collinear,

> (d) the Euler Lines of the three triangles are parallel,

> this is an easy consequence of (c).

>

> Of course the isogonal conjugates in (b) and (c) are the

> Isodynamic Points of their triangles. Thus, if F = X13,

> then the circle is that through X13, X14 and X16, and

> the line is X13X15. Note that F need not be X13 for the

> other two triangles, but is a Fermat Point.

>

> I suspect that (c) characterizes the Fermat Points, and

> perhaps that (d) also does.

>

> I'd be glad of any proofs or comments on these. I think

> that a circle with seven points is unusual. The centre

> might be interesting, it is on X115X125 as this is the

> perpendicular bisector of X13X14.

>

> Seasonal Greetings

>

> Wilson

>

>

>

>

>

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> - Dear Bernard,

I know nothing about angular coordinates, so hesitate to say much

Your proof below suggests that, if F = X13, then it is also X13 of

its cevian and anticevian triangle. This isn't always so, it can

be X14 for one or other.

> > We have the following fact :

coordinates

> >

> > If F is a Fermat Point of triangle ABC, then F is also a

> > Fermat Point of its Cevian, and of its Anticevian Triangle.

>

> this is direct consequence of the fact that the angular

> of a point wrt ABC, its cevian and anticevian triangles are the

same.

Is this true?

I'm guessing at the definition, but I think there may be

supplements involved. Probably I just guessed wrongly.

> for X13 : 120°, 120°, 120°

Yours

>

> [snip]

>

>

Wilson - Ah, that is what I wanted to know. But it seems to me that their

angular coordinates would not change, so both points should be the

same. It puzzles me that the second points are different. Intriguing

if they are.

Oh I know, it must be true that if we get the pedal and antipedal of

Fs, then the three Fs's are the same. This makes 12 points from these

six triangles, which I bet have an interesting relationship, Can't

wait until I have some time with Sketchpad.

Hmm, algebra at work.

Steve

On Dec 8, 2005, at 9:55 AM, Wilson Stothers wrote:

> the three triangles will have different second

> Fermat Points, so the lines FnFs are not parallel.

[Non-text portions of this message have been removed] - Bernard,

The two Fermat points each have a cevian and anticevian triangle. The

4 Euler lines of these triangles seem to be parallel to that of ABC

(observed on Sketchpad, not proved). Is there a curve, all of whose

points have Euler line parallel to that of ABC?

Steve - Dear Steve,

> The two Fermat points each have a cevian and anticevian triangle. The

It is true.

> 4 Euler lines of these triangles seem to be parallel to that of ABC

> (observed on Sketchpad, not proved).

> Is there a curve, all of whose

this was answered by Wilson yesterday under the thread [EMHL] The

> points have Euler line [of their cevian and anticevian triangles, I

> suppose] parallel to that of ABC?

Cubic K278!

Best regards

Bernard

[Non-text portions of this message have been removed]